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Rajeev Balasubramonian
Rajeev Balasubramonian
Rajeev Balasubramonian
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@user-bm4ig4fw1t
@user-bm4ig4fw1t 17 дней назад
very nice,..do y ouhave videos on mesi, moesi as well
@quercus_opuntia
@quercus_opuntia 2 месяца назад
this video saved my life and marraige
@quercus_opuntia
@quercus_opuntia 2 месяца назад
Thank u Rajeev!!!
@hussainfathy1065
@hussainfathy1065 6 месяцев назад
thank you <3
@VikramReddyAnapana
@VikramReddyAnapana 7 месяцев назад
Excellent teaching, and so well explained. I absorbed. Thank you so much
@VikramReddyAnapana
@VikramReddyAnapana 7 месяцев назад
Thank you so much.
@enzoding7558
@enzoding7558 7 месяцев назад
it should be -128~127 instead of -127~128
@lulu-xp7mf
@lulu-xp7mf 8 месяцев назад
thank you
@EriknocTDW
@EriknocTDW 11 месяцев назад
The 1st LW instruction has 8(R4) and the 2nd has 16(R4). What are the numbers 8 and 16 for?
@enzoding7558
@enzoding7558 7 месяцев назад
4 byte, 8 byte, 16 byte, etc. that's the offset from the stack pointer or from the current address of a register(e.g. R4, R3, R8, etc)
@shridharbendi9087
@shridharbendi9087 Год назад
Excellent presentation 👏👏
@rashedh2009
@rashedh2009 Год назад
Annoying voice
@achieverakash5192
@achieverakash5192 Год назад
you are super sir tomorrow is my exam you are my saviour
@chinmayrath8494
@chinmayrath8494 Год назад
Awesome explanation. Thank you !!!!
@raghul1208
@raghul1208 Год назад
best explanation
@MaxAbramson3
@MaxAbramson3 Год назад
Nowadays, compile time branch prediction (with profiling) is usually better than 95%, and run time branch prediction (one cycle beforehand) is about 97%, according to some manufacturers. So branching code actually suffers from this technique as it make the code larger, resulting in both higher ICache mispredict rate and pressure on the instruction bandwidth. It's not ironic that the only surviving major architectures are those without a delayed branch.
@MaxAbramson3
@MaxAbramson3 Год назад
Perfect branch prediction recognizes that only about 2-3% of branches are actually misbehaving, though we're only about to see about 90% of the time what direction a branch will go well ahead of time.
@ggxue
@ggxue Год назад
nice thank you
@archgirl2665
@archgirl2665 Год назад
Great video thankyou sir!
@chelseaethiohighlights1971
@chelseaethiohighlights1971 Год назад
Thank you! excuse me, would you provide slide for us?
@gabrieldias6430
@gabrieldias6430 Год назад
you are life savier, man
@ok-jg9jb
@ok-jg9jb Год назад
Tq sir
@zalatanibrah1803
@zalatanibrah1803 Год назад
concise explanation..
@rajeshhariharan7575
@rajeshhariharan7575 Год назад
Prof. Rajeev, Thanks a lot. very nicely explained.
@TheAyanMan
@TheAyanMan 2 года назад
Great vid ☺️
@TheAyanMan
@TheAyanMan 2 года назад
First comment 😃
@abhishekghosh1998
@abhishekghosh1998 2 года назад
Also the fast algorithm picture, (as given in Hennessy Patterson Textbook) seems a bit erroneous to me. When we are adding Multiplicand.Multiplier[0] + Multiplicand.Multiplier[1], then wouldn't we need to shift the Multiplicand.Multiplier[1] by 1 unit to the left for addition alignment? The fast multiplier implementation given in the Carl Hamacher textbook, explains it properly (where the entire hardware is given in details). Actually, Multiplicand.Multiplier[0], this gives a 32 bit number, the LSB of which forms product[0].... And the rest of the bits of Multiplicand.Multiplier[0], i.e. from 1 to 31st are given to the first level adder... such that Multiplicand.Multiplier[0][1] is aligned with Multiplicand.Multiplier[1][0]. Please correct me if I am wrong.
@abhishekghosh1998
@abhishekghosh1998 2 года назад
I do not think that the previous algorithm also works for signed numbers (I mean negative numbers) in all cases. The previous algorithm works in case of signed numbers, I guess, only when the multiplier is positive. The multiplicand can be positive or negative. But while right shifting the partial product formed at each step, we need to do an arithmetic right shift (instead of a simple logical right shift). Please correct me if I am wrong.
@souravgupta8182
@souravgupta8182 2 года назад
Awesome Video!!
@jaysiddhapura
@jaysiddhapura 2 года назад
Does DA and AD conversion needed between each layer !?
@weirdsciencetv4999
@weirdsciencetv4999 2 года назад
Why would we need to keep the resolution as opposed to just making a network which is tolerant of error accumulation?
@Max-ge7sv
@Max-ge7sv 2 года назад
If you have a network like this, the currents will not simply be added. In fact you have a complex current divider with multiple voltage sources. The output current is the result of the superposition of all current dividers, which depend on the resistor values and it will get more and more complex by increasing the input vector. Furthermore, the memrisors are changing their values by applying a voltage. How is it possible to get a consistent result?
@nabhay583
@nabhay583 Год назад
fWhat if we ground the lines? Won't we then easily be able to add the currents due to superposition?
@omersakkar5670
@omersakkar5670 2 года назад
أشكرك
@arnabsaha2021
@arnabsaha2021 2 года назад
Very well explained
@hungke6211
@hungke6211 2 года назад
how did you do it can you share with me , thank you
@warcroft23
@warcroft23 2 года назад
I came across this channel in 2015; I was pursuing my master's. I was impressed by your way of teaching and articulate explanations. Since then, I have been revisiting this channel whenever I need to recall Computer Organization concepts.
@sugee98
@sugee98 2 года назад
what happens is you've a heterogeneous multicore system with one processing element (say a DSP, for example) that has caches but doesn't participate in the MSI/MESI/MOESI protocol. I presume a read request will cause it to get he correct data from either main memory or another cache. But what if it wishes to modify that location and has, say write back cache). How do the other processing elements know it's been modified? Must the DSP do something in software to alert the other PEs? Excellent tutorial - the best I've seen in fact!
@manarzh7460
@manarzh7460 2 года назад
Thank you Rajeev
@eurotrash4970
@eurotrash4970 2 года назад
I have to say I love your british accent with a hint of indian as well. A lot less thick than most Indian computer science teachers you can find, and lot easier to understand
@pulkitjain25
@pulkitjain25 2 года назад
Also, what is the size of the the directory typically ?
@pulkitjain25
@pulkitjain25 2 года назад
Is the Directory on each node and is it kept coherent across all nodes? Or it is a in a global shared memory ?
@living_curious
@living_curious 2 года назад
How the directory is updated on all nodes?
@rajatbhattacharjya1443
@rajatbhattacharjya1443 3 года назад
This playlist needs to go viral
@vamosabv
@vamosabv 3 года назад
Thanks Rajeev for sharing :)
@akhilhooda740
@akhilhooda740 3 года назад
thanks sir ,you made easy
@saurabhdp
@saurabhdp 3 года назад
Prof. Rajeev, thanks for the video and your explanation. I have 1 question. If a situation arises where, say for example, both Processor P1 and P2 want to write to their copies of x exactly at the same time and issue an upgrade request simultaneously. Then how is this issue resolved. Thanks.
@bhavanibedre
@bhavanibedre 3 года назад
Your content is amazing. Do you have a sharable reference to your slides? It will be helpful to revise.
@ulasakyildiz
@ulasakyildiz 3 года назад
ADAM OĞLU ADAM Thank you brother
@MelvinSimon777
@MelvinSimon777 3 года назад
Amazingly explained Sir!
@vipulsharma3846
@vipulsharma3846 3 года назад
Nice explanation of clock speed
@naziyaakhtha3424
@naziyaakhtha3424 3 года назад
Very helpful
@ssvemuri
@ssvemuri 3 года назад
Nice to see you on youtube Rajeev. Hope you are doing well!
@padalatejasaikumarreddy5321
@padalatejasaikumarreddy5321 3 года назад
wowwwwwwwwwwwwww